Trong không gian $\text{Ox}yz$, cho các điểm...

$\overrightarrow{AB}=\left( 1;0;0 \right);\overrightarrow{AC}=\left( 0;0;2 \right);Cos(\overrightarrow{AB};\overrightarrow{AC})=0\Rightarrow $ $\widehat{BAC}={{90}^{0}}$
$\dfrac{\overrightarrow{AB}}{\left| \overrightarrow{AB} \right|}=\left( 1;0;0 \right);\dfrac{\overrightarrow{AC}}{\left| \overrightarrow{AC} \right|}=\left( 0;0;1 \right)$
Suy ra: Có hai tia phân giác là ${{d}_{1}},{{d}_{2}}$
$\overrightarrow{{{u}_{{{d}_{1}}}}}=\dfrac{\overrightarrow{AB}}{\left| \overrightarrow{AB} \right|}+\dfrac{\overrightarrow{AC}}{\left| \overrightarrow{AC} \right|}=\left( 1;0;1 \right)$
$\overrightarrow{{{u}_{{{d}_{2}}}}}=\dfrac{\overrightarrow{AB}}{\left| \overrightarrow{AB} \right|}-\dfrac{\overrightarrow{AC}}{\left| \overrightarrow{AC} \right|}=\left( 1;0;-1 \right)$
$\begin{aligned}
& \Rightarrow ({{d}_{1}}):\left\{ \begin{aligned}
& x=1+t \\
& y=1 \\
& z=2+t \\
\end{aligned} \right. \\
& ({{d}_{2}}):\left\{ \begin{aligned}
& x=1+t \\
& y=1 \\
& z=2-t \\
\end{aligned} \right. \\
\end{aligned}$
$\begin{aligned}
& (P):x+y=0 \\
& \Rightarrow 1+t+1=0 \\
& \Rightarrow t=-2 \\
& \Rightarrow M(-1;1;0) \\
& \Rightarrow a+b=0 \\
\end{aligned}$
Cách 2:

Gọi I là chân đường phân giác
$\Rightarrow \dfrac{IC}{IB}=\dfrac{AC}{AB}=2\Rightarrow IC=2IB$
$\overrightarrow{IC}=-2{{\overrightarrow{IB}}^{{}}}(*)\Rightarrow I\left( \dfrac{5}{3};1;\dfrac{8}{3} \right)$
$\Rightarrow \overrightarrow{AI}=(1;0;1)\Rightarrow (AI):\left\{ \begin{aligned}
& x=1+t \\
& y=1 \\
& z=2+t \\
\end{aligned} \right.$
$\begin{aligned}
& (Oxy):z=0 \\
& \Rightarrow \left\{ M \right\}=AI\cap (Oxy) \\
& \Rightarrow t=-2 \\
& \Rightarrow M(-1;1;0) \\
& \Rightarrow a+b=0 \\
\end{aligned}$